I am looking over a past paper for which I do not have the mark scheme for and the question asks to evaluate the following contour integrals: $$\int_{|z|=2} |z+1|^2dz$$ $$\int_{|z|=2} (z+1)^2|dz|$$ I can't see a way to use Cauchy's integral formula and I am unsure how I would go about this using a substitution, help greatly appreciated.

  • $\begingroup$ what does the integral with $|dz|$ even mean $\endgroup$ – amakelov Apr 21 '17 at 2:33

I don't think you've got any choice but to parametrise the curve, short of knowing a few tricks. For example, the first one is $$ \int_{|z|=2} (\lvert z \rvert^2 + z + \bar{z} + 1) \, dz. $$ The last term, $1$, is analytic, so integrates to zero on a closed contour by Cauchy's theorem. $z$ is likewise analytic and so integrates to zero. $\lvert z \rvert^2$ is constant on the contour, equal to $4$, so the integral is zero again because a constant is analytic. Lastly, in the same way $$ \bar{z} = \frac{4}{z}, $$ so the only thing left in the integral is $$ 4\int_{|z|=2} \frac{dz}{z} = 4 \cdot 2\pi i $$ by the residue theorem (or the Cauchy integral formula, if you prefer). You might notice this is $2i$ times the area inside the circle of radius $2$; a nice thing about $\oint_{\gamma} \bar{z} \, dz$ for simple $\gamma$ is that it gives you $2i$ times the area inside $\gamma$; this is quite easy to prove using Green's theorem.

So the first one was possible to do without parametrising the contour, but I don't think this is possible for the second one: $\lvert dz \rvert = 2 \, d\theta$, where $0\leq\theta<2\pi$ and that's basically the only way to go: $\lvert dz \rvert$ is arc-length, so it's an arc-length integral, not a complex one. We have $$ (1+z)^2 = (1+2e^{i\theta})^2 = 1+4e^{i\theta}+4e^{2i\theta}, $$ and of course the second two terms integrate to zero over $[0,2\pi)$, so the integral is $1 \cdot \int_0^{2\pi} 2d\theta = 4\pi$.

  • $\begingroup$ Great answer, really helped me to understand how to answer these sorts of questions, +1. Just one question, what do I do if integrating over the circle $|z-1|=1$ for example? Is it the same thing just different singularities or would the method need to be changed? $\endgroup$ – Matt Apr 21 '17 at 2:49
  • $\begingroup$ If it's a circle with a different centre, a change of variables will suffice to move it to be centred at the origin and the same method applies. Or one can use $\bar{z}-1 = \frac{1}{z-1}$ and so on: it'll come out the same in the end. $\endgroup$ – Chappers Apr 21 '17 at 3:02

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